Prove that zbar z = |z|^2

Prove that $z\bar z = |z|^2$.

Step 1 · Set up $z$ and $\bar z$

Let $z = x+iy$ with $x,y\in\mathbb R$. Its conjugate flips the sign of the imaginary part:

$$ \bar z = x-iy $$

Step 2 · Compute $z\bar z$

Using $(a+b)(a-b)=a^2-b^2$ with $a=x$, $b=iy$:

$$ z\bar z = (x+iy)(x-iy) = x^2-(iy)^2 = x^2-i^2y^2 = x^2+y^2 $$

Step 3 · Compute $|z|^2$

By definition $|z|=\sqrt{x^2+y^2}$, so:

$$ |z|^2 = x^2+y^2 $$

Step 4 · Compare

Both computations give the same real number:

$$ \boxed{z\bar z = x^2+y^2 = |z|^2} $$

$\blacksquare$

Step 5 · Cross-check with examples

$z=3+4i$: $z\bar z=(3+4i)(3-4i)=9-16i^2=9+16=25$, and $|z|^2=(\sqrt{9+16})^2=25$. ✓

$z=1+i$: $z\bar z=(1+i)(1-i)=1-i^2=2$, and $|z|^2=(\sqrt2)^2=2$. ✓

$z=-2+3i$: $z\bar z=(-2+3i)(-2-3i)=4-9i^2=13$, and $|z|^2=(\sqrt{13})^2=13$. ✓

Step 6 · Independent proof via polar form

Let $z = re^{i\theta}$ with $r=|z|$. Then $\bar z = re^{-i\theta}$, so:

$$ z\bar z = re^{i\theta}\cdot re^{-i\theta} = r^2 e^{i(\theta-\theta)} = r^2 e^0 = r^2 = |z|^2 $$

confirming the identity independently of the Cartesian derivation.

Step 7 · Why this identity matters

Finding the modulus: $|z| = \sqrt{z\bar z}$.

Division by a complex number: $\dfrac1z = \dfrac{\bar z}{z\bar z} = \dfrac{\bar z}{|z|^2}$ — this is exactly how division by complex numbers is carried out in practice.

Always real and non-negative: since $z\bar z = x^2+y^2 \ge 0$, and $z\bar z = 0$ exactly when $z=0$.

Geometric meaning: $|z|^2$ is the squared distance from the origin to $z$ in the complex plane; $z\bar z$ gives an algebraic way to compute that squared distance without ever taking a square root.